Primitive of x over a x + b by p x + q/Partial Fraction Expansion

Lemma for Primitive of x over a x + b by p x + q

$\dfrac x {\paren {a x + b} \paren {p x + q} } \equiv \dfrac b {\paren {b p - a q} \paren {a x + b} } - \dfrac q {\paren {b p - a q} \paren {p x + q} }$

Proof

\(\ds \frac x {\paren {a x + b} \paren {p x + q} }\)

\(\equiv\)

\(\ds \frac A {a x + b} + \frac B {p x + q}\)

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\equiv\)

\(\ds A \paren {p x + q} + B \paren {a x + b}\)

multiplying through by $\paren {a x + b} \paren {p x + q}$

Setting $p x + q = 0$ in $(1)$:

\(\ds p x + q\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds -\frac q p\)

\(\ds \leadsto \ \ \)

\(\ds B \paren {a \paren {-\frac q p} + b}\)

\(=\)

\(\ds -\frac q p\)

substituting for $x$ in $(1)$: term in $p x + q$ is $0$

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds \frac {-q} {b p - a q}\)

Setting $a x + b = 0$ in $(1)$:

\(\ds a x + b\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds -\frac b a\)

\(\ds \leadsto \ \ \)

\(\ds A \paren {p \paren {-\frac b a} + q}\)

\(=\)

\(\ds -\frac b a\)

substituting for $x$ in $(1)$: term in $a x + b$ is $0$

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds \frac b {b p - a q}\)

Summarising:

\(\ds A\)

\(=\)

\(\ds \frac b {p b - a q}\)

\(\ds B\)

\(=\)

\(\ds \frac {-q} {b p - a q}\)

Hence the result.

$\blacksquare$